Basic force with static friction problem

A 70-kg skier is in a tuck position and moving down a 20-degree hill. Air resistance applies a resultant force of 15 N against the movement of the skier. The coefficient of friction between the skis and the snow is 0.09. What is the resultant downhill force acting on the skier?

2. Relevant equations

Force = friction coefficient x normal force

3. The attempt at a solution

I know you have to break down the force into its x and y components. So making a right triangle with the weight (~700 N: 70kg x 10 m/s2) as the hypotenuse and 70 degrees (90 - the given 20 degrees) as the angle. But after doing that, I'm not sure where to go from there. Is the resultant force also your x-component in the right triangle?